Difference between revisions of "2017 AMC 12B Problems/Problem 18"

(Solution 4 (Coordinate Geometry))
(Solution 5 (No sqrts))
 
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We can set AF = 49, CF = 35, BF = 25 and scale back later
 
We can set AF = 49, CF = 35, BF = 25 and scale back later
  
Then the radius is AB/2 = (AF+BF)/2 = 74/2 = 37.  
+
Then the radius is <math>\frac{AB}{2}</math> = <math>\frac{AF+BF}{2}</math> = <math>\frac{74}{2}</math> = <math>37</math>.  
  
 
So the radius is 37 and the height of ABC is 35.
 
So the radius is 37 and the height of ABC is 35.
  
If we scale it back so that our radius is 2, our height is 70/37.
+
If we scale it back so that our radius is 2, our height is <math>\frac{70}{37}</math>.
  
 
Area of ABC is bh/2 = <math>\frac{(4)(\frac{70}{37})}{2}</math> = <math>\boxed{\textbf{(D)}\ \frac{140}{37}}</math>.
 
Area of ABC is bh/2 = <math>\frac{(4)(\frac{70}{37})}{2}</math> = <math>\boxed{\textbf{(D)}\ \frac{140}{37}}</math>.

Latest revision as of 21:44, 30 September 2024

Problem

The diameter $AB$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$. Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$. Segment $AE$ intersects the circle at a point $C$ between $A$ and $E$. What is the area of $\triangle  ABC$?

$\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}$


Solution 1

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Let $O$ be the center of the circle. Note that $EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}$. However, by Power of a Point, $(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}$, so $AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}$. Now $BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}$. Since $\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}$.


Solution 2: Similar triangles with Pythagorean

$AB$ is the diameter of the circle, so $\angle ACB$ is a right angle, and therefore by AA similarity, $\triangle ACB \sim \triangle ADE$.

Because of this, $\frac{AC}{AD} = \frac{AB}{AE} \Longrightarrow \frac{AC}{2+2+3} = \frac{2+2}{\sqrt{7^2 + 5^2}}$, so $AC = \frac{28}{\sqrt{74}}$.

Likewise, $\frac{BC}{ED} = \frac{AB}{AE} \Longrightarrow \frac{BC}{5} = \frac{4}{\sqrt{74}}$, so $BC = \frac{20}{\sqrt{74}}$.

Thus the area of $\triangle ABC = \frac{1}{2} \cdot \frac{28}{\sqrt{74}} \cdot \frac{20}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}$.

Solution 2b: Area shortcut

Because $AE$ is $\sqrt{74}$ and $AB$ is $4$, the ratio of the sides is $\frac{\sqrt{74}}{4}$, meaning the ratio of the areas is thus ${(\frac{\sqrt{74}}{4})}^2 \implies \frac{74}{16} \implies \frac{37}{8}$. We then have the proportion $\frac{\frac{5*7}{2}}{[ABC]}=\frac{37}{8} \implies 37*[ABC]=140 \implies \boxed{\textbf{(D)}\ \frac{140}{37}}$

Solution 3: Similar triangles without Pythagorean

Or, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:

Draw $BF \parallel ED$ with $F$ on $AE$. $BF=5\times\frac{4}{7}=\frac{20}{7}$.

$[\triangle ABF]=\frac{1}{2} \times 4 \times \frac{20}{7}=\frac{40}{7}$.

$AC:CB:CF=49:35:25$. ($7:5$ ratio applied twice)

$[\triangle ABC]=\frac{49}{49+25}[\triangle ABF]=\boxed{\textbf{(D)}\ \frac{140}{37}}$.

Solution 4 (Coordinate Geometry)

Let $A$ be at the origin $(0, 0)$ of a coordinate plane, with $B$ being located at $(4, 0)$, etc.

We can find the area of $\triangle ABC$ by finding the the altitude from line $AB$ to point $C$. Realize that this altitude is the $y$ coordinate of point $C$ on the coordinate plane, since the respective base of $\triangle ABC$ is on the $x$-axis.

Using the diagram in solution one, the equation for circle $O$ is $(x-2)^2+y^2 = 4$.

The equation for line $AE$ is then $y = \frac{5}{7}x$, therefore $x = \frac{7}{5}y$.

Substituting $\frac{7}{5}y$ for $x$ in the equation for circle $O$, we get:

$\left(\frac{7}{5}y-2\right)^2+y^2 = 4$

We can solve for $y$ to yield the $y$ coordinate of point $C$ in the coordinate plane, since this is the point of intersection of the circle and line $AE$. Note that one root will yield the intersection of the circle and line $AE$ at the origin, so we will ignore this root.

Expanding the expression and factoring, we get:

$\left(\frac{49}{25}y^2-\frac{28}{5}y+4\right)+y^2 = 4$

$\frac{74}{25}y^2-\frac{28}{5}y = 0$

$50y(37y-70) = 0$

Our non-zero root is thus $\frac{70}{37}$. Calculating the area of $\triangle ABC$ with $4$ as the length of $AB$ and $\frac{70}{37}$ as the altitude, we get:

$\frac{(4)(\frac{70}{37})}{2} = \boxed{\textbf{(D)}\ \frac{140}{37}}$.

-Solution by Joeya

Solution 5 (No sqrts)

Slope of AC is 5/7 As stated in other solutions AB is the diameter, ABC is right.

Let CF be an altitude of ABC.

AF:CF = CF:BF = 7:5

We can set AF = 49, CF = 35, BF = 25 and scale back later

Then the radius is $\frac{AB}{2}$ = $\frac{AF+BF}{2}$ = $\frac{74}{2}$ = $37$.

So the radius is 37 and the height of ABC is 35.

If we scale it back so that our radius is 2, our height is $\frac{70}{37}$.

Area of ABC is bh/2 = $\frac{(4)(\frac{70}{37})}{2}$ = $\boxed{\textbf{(D)}\ \frac{140}{37}}$.

-mathophobia

Video Solution by OmegaLearn (Similar Triangles)

https://youtu.be/NsQbhYfGh1Q?t=512

~ pi_is_3.14

See Also

2017 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 12 Problems and Solutions

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